Mixing
Iterated function 'periodicity'
Note: $f^n$ denotes the iteration of composition, e.g. $f^3(x)=(fcirc fcirc f)(x)$
I've noticed that particular functions have a certain property where for some number $n$ the iterations of the function cycle through a set of values so that $f^{m+n}(x)=f^m(x)$ for all $minmathbb{N}$. For example, if $f(x)=1-frac{1}{x}$:
$$f^1(x)=1-frac{1}{x}$$
$$f^2(x)=1-frac{1}{1-frac{1}{x}}=frac{1}{1-x}$$
$$f^3(x)=1-frac{1}{1-frac{1}{1-frac{1}{x}}}=x$$
$$f^4(x)=1-frac{1}{x}$$
So the cycle has a period of $n=3$. Is there a name for this property, and where can I find more information? Also, are there any cases where the 'period' $n$ varies as a function of the iterate $m$?
Edit:
As others have pointed to in the comments, the property I am describing can be stated succinctly by $F^n(X)=X$* for some $n$, and can apply to functions as well as operators on functions.
Since this extends rather naturally to integer $n$, idempotence and involution would be examples with periods $1$ and $2$, respectively.
If matrix multiplication is used to represent the composition of functions, then the property in question applies to any $M$ such that $M^n=pm I$ for some $n$. As Will Jagy pointed out, in the example $f(x)=1-frac{1}{x}$, $M^3=-I$ is given by the Moebius transformation $f(x)=frac{x-1}{x+0}$.
Given how incredibly general this property is and the number of things to which it applies there is absolutely no way that I am the first person to notice it. There has to be a book or a paper somewhere, right?
*In retrospect, this should have been apparent given that $f^{m+n}=f^mimplies f^n=f^0$
"Corollary"?
If $$frac{d^nf(x)}{dx^n}=f(x)$$ for some $ninmathbb{Z},nneq0$, then $$frac{d^{mn}f(x)}{dx^{mn}}=f(x)$$ and $$int^{m(n-1)}f(x) dx^{m(n-1)}=f(x)$$ for all $minmathbb{Z}$
abstract-algebra polynomials function-and-relation-composition
asked Dec 31 '18 at 23:08
R. BurtonR. Burton
36919
add a comment |
Note: $f^n$ denotes the iteration of composition, e.g. $f^3(x)=(fcirc fcirc f)(x)$
I've noticed that particular functions have a certain property where for some number $n$ the iterations of the function cycle through a set of values so that $f^{m+n}(x)=f^m(x)$ for all $minmathbb{N}$. For example, if $f(x)=1-frac{1}{x}$:
$$f^1(x)=1-frac{1}{x}$$
$$f^2(x)=1-frac{1}{1-frac{1}{x}}=frac{1}{1-x}$$
$$f^3(x)=1-frac{1}{1-frac{1}{1-frac{1}{x}}}=x$$
$$f^4(x)=1-frac{1}{x}$$
So the cycle has a period of $n=3$. Is there a name for this property, and where can I find more information? Also, are there any cases where the 'period' $n$ varies as a function of the iterate $m$?
Edit:
As others have pointed to in the comments, the property I am describing can be stated succinctly by $F^n(X)=X$* for some $n$, and can apply to functions as well as operators on functions.
Since this extends rather naturally to integer $n$, idempotence and involution would be examples with periods $1$ and $2$, respectively.
If matrix multiplication is used to represent the composition of functions, then the property in question applies to any $M$ such that $M^n=pm I$ for some $n$. As Will Jagy pointed out, in the example $f(x)=1-frac{1}{x}$, $M^3=-I$ is given by the Moebius transformation $f(x)=frac{x-1}{x+0}$.
Given how incredibly general this property is and the number of things to which it applies there is absolutely no way that I am the first person to notice it. There has to be a book or a paper somewhere, right?
*In retrospect, this should have been apparent given that $f^{m+n}=f^mimplies f^n=f^0$
"Corollary"?
If $$frac{d^nf(x)}{dx^n}=f(x)$$ for some $ninmathbb{Z},nneq0$, then $$frac{d^{mn}f(x)}{dx^{mn}}=f(x)$$ and $$int^{m(n-1)}f(x) dx^{m(n-1)}=f(x)$$ for all $minmathbb{Z}$
abstract-algebra polynomials function-and-relation-composition
asked Dec 31 '18 at 23:08
R. BurtonR. Burton
36919
1
You can think of $f(x)$ as the position of a particle at time $n+1$ when the position at time $n$ is $x$. Then $f^{n}(x)=x$ says the $x$ is a periodic point with $n$ as a period.
– Kavi Rama Murthy
Dec 31 '18 at 23:22
If, for a positive integer $k$, $f^k(x)=x$ for all $x$ in the domain of $f$, then $f$ is $I^{1/k}$, a fractional iterate of the identity function.
– r.e.s.
Jan 1 at 4:13
@KaviRamaMurthy Now THAT's an interesting point. I hadn't considered $f^n(x)=f^0(x)$ to be a criterion for periodicity, but in retrospect it makes sense. In fact, this seems to work for all integer iterates, since $f^{-1}(x)=frac{1}{1-x}$ and $f^{-2}(x)=1-frac{1}{x}$. $f$ doesn't even need to be a function, since this could apply to any operator for which $f^0$ is defined.
– R. Burton
Jan 3 at 18:53
add a comment |
Note: $f^n$ denotes the iteration of composition, e.g. $f^3(x)=(fcirc fcirc f)(x)$
I've noticed that particular functions have a certain property where for some number $n$ the iterations of the function cycle through a set of values so that $f^{m+n}(x)=f^m(x)$ for all $minmathbb{N}$. For example, if $f(x)=1-frac{1}{x}$:
$$f^1(x)=1-frac{1}{x}$$
$$f^2(x)=1-frac{1}{1-frac{1}{x}}=frac{1}{1-x}$$
$$f^3(x)=1-frac{1}{1-frac{1}{1-frac{1}{x}}}=x$$
$$f^4(x)=1-frac{1}{x}$$
So the cycle has a period of $n=3$. Is there a name for this property, and where can I find more information? Also, are there any cases where the 'period' $n$ varies as a function of the iterate $m$?
Edit:
As others have pointed to in the comments, the property I am describing can be stated succinctly by $F^n(X)=X$* for some $n$, and can apply to functions as well as operators on functions.
Since this extends rather naturally to integer $n$, idempotence and involution would be examples with periods $1$ and $2$, respectively.
If matrix multiplication is used to represent the composition of functions, then the property in question applies to any $M$ such that $M^n=pm I$ for some $n$. As Will Jagy pointed out, in the example $f(x)=1-frac{1}{x}$, $M^3=-I$ is given by the Moebius transformation $f(x)=frac{x-1}{x+0}$.
Given how incredibly general this property is and the number of things to which it applies there is absolutely no way that I am the first person to notice it. There has to be a book or a paper somewhere, right?
*In retrospect, this should have been apparent given that $f^{m+n}=f^mimplies f^n=f^0$
"Corollary"?
If $$frac{d^nf(x)}{dx^n}=f(x)$$ for some $ninmathbb{Z},nneq0$, then $$frac{d^{mn}f(x)}{dx^{mn}}=f(x)$$ and $$int^{m(n-1)}f(x) dx^{m(n-1)}=f(x)$$ for all $minmathbb{Z}$
abstract-algebra polynomials function-and-relation-composition
asked Dec 31 '18 at 23:08
R. BurtonR. Burton
36919
Note: $f^n$ denotes the iteration of composition, e.g. $f^3(x)=(fcirc fcirc f)(x)$
I've noticed that particular functions have a certain property where for some number $n$ the iterations of the function cycle through a set of values so that $f^{m+n}(x)=f^m(x)$ for all $minmathbb{N}$. For example, if $f(x)=1-frac{1}{x}$:
$$f^1(x)=1-frac{1}{x}$$
$$f^2(x)=1-frac{1}{1-frac{1}{x}}=frac{1}{1-x}$$
$$f^3(x)=1-frac{1}{1-frac{1}{1-frac{1}{x}}}=x$$
$$f^4(x)=1-frac{1}{x}$$
So the cycle has a period of $n=3$. Is there a name for this property, and where can I find more information? Also, are there any cases where the 'period' $n$ varies as a function of the iterate $m$?
Edit:
As others have pointed to in the comments, the property I am describing can be stated succinctly by $F^n(X)=X$* for some $n$, and can apply to functions as well as operators on functions.
Since this extends rather naturally to integer $n$, idempotence and involution would be examples with periods $1$ and $2$, respectively.
If matrix multiplication is used to represent the composition of functions, then the property in question applies to any $M$ such that $M^n=pm I$ for some $n$. As Will Jagy pointed out, in the example $f(x)=1-frac{1}{x}$, $M^3=-I$ is given by the Moebius transformation $f(x)=frac{x-1}{x+0}$.
Given how incredibly general this property is and the number of things to which it applies there is absolutely no way that I am the first person to notice it. There has to be a book or a paper somewhere, right?
*In retrospect, this should have been apparent given that $f^{m+n}=f^mimplies f^n=f^0$
"Corollary"?
If $$frac{d^nf(x)}{dx^n}=f(x)$$ for some $ninmathbb{Z},nneq0$, then $$frac{d^{mn}f(x)}{dx^{mn}}=f(x)$$ and $$int^{m(n-1)}f(x) dx^{m(n-1)}=f(x)$$ for all $minmathbb{Z}$
abstract-algebra polynomials function-and-relation-composition
abstract-algebra polynomials function-and-relation-composition
asked Dec 31 '18 at 23:08
R. BurtonR. Burton
36919
asked Dec 31 '18 at 23:08
R. BurtonR. Burton
36919
edited Jan 3 at 19:49
R. Burton
asked Dec 31 '18 at 23:08
R. BurtonR. Burton
36919
asked Dec 31 '18 at 23:08
R. BurtonR. Burton
36919
asked Dec 31 '18 at 23:08
R. BurtonR. Burton
36919
36919
1
You can think of $f(x)$ as the position of a particle at time $n+1$ when the position at time $n$ is $x$. Then $f^{n}(x)=x$ says the $x$ is a periodic point with $n$ as a period.
– Kavi Rama Murthy
Dec 31 '18 at 23:22
If, for a positive integer $k$, $f^k(x)=x$ for all $x$ in the domain of $f$, then $f$ is $I^{1/k}$, a fractional iterate of the identity function.
– r.e.s.
Jan 1 at 4:13
@KaviRamaMurthy Now THAT's an interesting point. I hadn't considered $f^n(x)=f^0(x)$ to be a criterion for periodicity, but in retrospect it makes sense. In fact, this seems to work for all integer iterates, since $f^{-1}(x)=frac{1}{1-x}$ and $f^{-2}(x)=1-frac{1}{x}$. $f$ doesn't even need to be a function, since this could apply to any operator for which $f^0$ is defined.
– R. Burton
Jan 3 at 18:53
add a comment |
1
You can think of $f(x)$ as the position of a particle at time $n+1$ when the position at time $n$ is $x$. Then $f^{n}(x)=x$ says the $x$ is a periodic point with $n$ as a period.
– Kavi Rama Murthy
Dec 31 '18 at 23:22
If, for a positive integer $k$, $f^k(x)=x$ for all $x$ in the domain of $f$, then $f$ is $I^{1/k}$, a fractional iterate of the identity function.
– r.e.s.
Jan 1 at 4:13
@KaviRamaMurthy Now THAT's an interesting point. I hadn't considered $f^n(x)=f^0(x)$ to be a criterion for periodicity, but in retrospect it makes sense. In fact, this seems to work for all integer iterates, since $f^{-1}(x)=frac{1}{1-x}$ and $f^{-2}(x)=1-frac{1}{x}$. $f$ doesn't even need to be a function, since this could apply to any operator for which $f^0$ is defined.
– R. Burton
Jan 3 at 18:53
1
1
You can think of $f(x)$ as the position of a particle at time $n+1$ when the position at time $n$ is $x$. Then $f^{n}(x)=x$ says the $x$ is a periodic point with $n$ as a period.
– Kavi Rama Murthy
Dec 31 '18 at 23:22
You can think of $f(x)$ as the position of a particle at time $n+1$ when the position at time $n$ is $x$. Then $f^{n}(x)=x$ says the $x$ is a periodic point with $n$ as a period.
– Kavi Rama Murthy
Dec 31 '18 at 23:22
If, for a positive integer $k$, $f^k(x)=x$ for all $x$ in the domain of $f$, then $f$ is $I^{1/k}$, a fractional iterate of the identity function.
– r.e.s.
Jan 1 at 4:13
If, for a positive integer $k$, $f^k(x)=x$ for all $x$ in the domain of $f$, then $f$ is $I^{1/k}$, a fractional iterate of the identity function.
– r.e.s.
Jan 1 at 4:13
@KaviRamaMurthy Now THAT's an interesting point. I hadn't considered $f^n(x)=f^0(x)$ to be a criterion for periodicity, but in retrospect it makes sense. In fact, this seems to work for all integer iterates, since $f^{-1}(x)=frac{1}{1-x}$ and $f^{-2}(x)=1-frac{1}{x}$. $f$ doesn't even need to be a function, since this could apply to any operator for which $f^0$ is defined.
– R. Burton
Jan 3 at 18:53
@KaviRamaMurthy Now THAT's an interesting point. I hadn't considered $f^n(x)=f^0(x)$ to be a criterion for periodicity, but in retrospect it makes sense. In fact, this seems to work for all integer iterates, since $f^{-1}(x)=frac{1}{1-x}$ and $f^{-2}(x)=1-frac{1}{x}$. $f$ doesn't even need to be a function, since this could apply to any operator for which $f^0$ is defined.
– R. Burton
Jan 3 at 18:53
add a comment |
1 Answer
1
active
oldest
votes
writing your function as the Moebius transformation
$$ frac{x-1}{x+0} $$
we get the matrix
$$
M =
left(
begin{array}{rr}
1 & -1 \
1 & 0
end{array}
right)
$$
Note that $M^2 -M+I=0$ so $M^3 = -I$
You can do something similar with any
$$
M =
left(
begin{array}{rr}
a & b \
c & d
end{array}
right)
$$
and
$$ frac{ax+b}{cx+d} $$
such that $M^n = I$ for some $n.$ It is also allowed to have the elements of $M$ complex if you want.
answered Jan 1 at 0:57
Will JagyWill Jagy
102k5100199
Okay, I had to learn what a Moebius transformation was first, but this only seems to apply to the quotients of linear functions, whereas functions with the property $f^n(x)=I(x)$, for some $n$, may not always be linear. Also shouldn't it be $M^2-M+I=0$ and $M^3=-I$?
– R. Burton
Jan 3 at 18:28
add a comment |
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writing your function as the Moebius transformation
$$ frac{x-1}{x+0} $$
we get the matrix
$$
M =
left(
begin{array}{rr}
1 & -1 \
1 & 0
end{array}
right)
$$
Note that $M^2 -M+I=0$ so $M^3 = -I$
You can do something similar with any
$$
M =
left(
begin{array}{rr}
a & b \
c & d
end{array}
right)
$$
and
$$ frac{ax+b}{cx+d} $$
such that $M^n = I$ for some $n.$ It is also allowed to have the elements of $M$ complex if you want.
answered Jan 1 at 0:57
Will JagyWill Jagy
102k5100199
Okay, I had to learn what a Moebius transformation was first, but this only seems to apply to the quotients of linear functions, whereas functions with the property $f^n(x)=I(x)$, for some $n$, may not always be linear. Also shouldn't it be $M^2-M+I=0$ and $M^3=-I$?
– R. Burton
Jan 3 at 18:28
add a comment |
writing your function as the Moebius transformation
$$ frac{x-1}{x+0} $$
we get the matrix
$$
M =
left(
begin{array}{rr}
1 & -1 \
1 & 0
end{array}
right)
$$
Note that $M^2 -M+I=0$ so $M^3 = -I$
You can do something similar with any
$$
M =
left(
begin{array}{rr}
a & b \
c & d
end{array}
right)
$$
and
$$ frac{ax+b}{cx+d} $$
such that $M^n = I$ for some $n.$ It is also allowed to have the elements of $M$ complex if you want.
answered Jan 1 at 0:57
Will JagyWill Jagy
102k5100199
Okay, I had to learn what a Moebius transformation was first, but this only seems to apply to the quotients of linear functions, whereas functions with the property $f^n(x)=I(x)$, for some $n$, may not always be linear. Also shouldn't it be $M^2-M+I=0$ and $M^3=-I$?
– R. Burton
Jan 3 at 18:28
add a comment |
writing your function as the Moebius transformation
$$ frac{x-1}{x+0} $$
we get the matrix
$$
M =
left(
begin{array}{rr}
1 & -1 \
1 & 0
end{array}
right)
$$
Note that $M^2 -M+I=0$ so $M^3 = -I$
You can do something similar with any
$$
M =
left(
begin{array}{rr}
a & b \
c & d
end{array}
right)
$$
and
$$ frac{ax+b}{cx+d} $$
such that $M^n = I$ for some $n.$ It is also allowed to have the elements of $M$ complex if you want.
answered Jan 1 at 0:57
Will JagyWill Jagy
102k5100199
writing your function as the Moebius transformation
$$ frac{x-1}{x+0} $$
we get the matrix
$$
M =
left(
begin{array}{rr}
1 & -1 \
1 & 0
end{array}
right)
$$
Note that $M^2 -M+I=0$ so $M^3 = -I$
You can do something similar with any
$$
M =
left(
begin{array}{rr}
a & b \
c & d
end{array}
right)
$$
and
$$ frac{ax+b}{cx+d} $$
such that $M^n = I$ for some $n.$ It is also allowed to have the elements of $M$ complex if you want.
answered Jan 1 at 0:57
Will JagyWill Jagy
102k5100199
edited Jan 3 at 18:30
answered Jan 1 at 0:57
Will JagyWill Jagy
102k5100199
answered Jan 1 at 0:57
Will JagyWill Jagy
102k5100199
answered Jan 1 at 0:57
Will JagyWill Jagy
102k5100199
102k5100199
Okay, I had to learn what a Moebius transformation was first, but this only seems to apply to the quotients of linear functions, whereas functions with the property $f^n(x)=I(x)$, for some $n$, may not always be linear. Also shouldn't it be $M^2-M+I=0$ and $M^3=-I$?
– R. Burton
Jan 3 at 18:28
add a comment |
Okay, I had to learn what a Moebius transformation was first, but this only seems to apply to the quotients of linear functions, whereas functions with the property $f^n(x)=I(x)$, for some $n$, may not always be linear. Also shouldn't it be $M^2-M+I=0$ and $M^3=-I$?
– R. Burton
Jan 3 at 18:28
Okay, I had to learn what a Moebius transformation was first, but this only seems to apply to the quotients of linear functions, whereas functions with the property $f^n(x)=I(x)$, for some $n$, may not always be linear. Also shouldn't it be $M^2-M+I=0$ and $M^3=-I$?
– R. Burton
Jan 3 at 18:28
Okay, I had to learn what a Moebius transformation was first, but this only seems to apply to the quotients of linear functions, whereas functions with the property $f^n(x)=I(x)$, for some $n$, may not always be linear. Also shouldn't it be $M^2-M+I=0$ and $M^3=-I$?
– R. Burton
Jan 3 at 18:28
add a comment |
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1
You can think of $f(x)$ as the position of a particle at time $n+1$ when the position at time $n$ is $x$. Then $f^{n}(x)=x$ says the $x$ is a periodic point with $n$ as a period.
– Kavi Rama Murthy
Dec 31 '18 at 23:22
If, for a positive integer $k$, $f^k(x)=x$ for all $x$ in the domain of $f$, then $f$ is $I^{1/k}$, a fractional iterate of the identity function.
– r.e.s.
Jan 1 at 4:13
@KaviRamaMurthy Now THAT's an interesting point. I hadn't considered $f^n(x)=f^0(x)$ to be a criterion for periodicity, but in retrospect it makes sense. In fact, this seems to work for all integer iterates, since $f^{-1}(x)=frac{1}{1-x}$ and $f^{-2}(x)=1-frac{1}{x}$. $f$ doesn't even need to be a function, since this could apply to any operator for which $f^0$ is defined.
– R. Burton
Jan 3 at 18:53